Question

Consider a medium with parameters € = 1.2 (10^-10 )F/m , n= 3(10^-3) H/m and sigma=0. Magnetic field intensity in the medium is given as R = 2cos (10^10t- 600x)äz Am.

Use Maxwell's equations to obtain the followings:
1) Magnetic flux density
These questions is circuit theory

Answer 1

Using Maxwell's equations, we can determine the magnetic flux density. One of the Maxwell's equations is:

`\displaystyle \\abla * \mathbf{H} = \mathbf{J} + \frac{\partial \mathbf{D}}{\partial t}`,

where
`\displaystyle \\abla * \mathbf{H}` is the curl of the magnetic field intensity
`\displaystyle \mathbf{H}`,
`\displaystyle \mathbf{J}` is the current density, and
`\displaystyle \frac{\partial \mathbf{D}}{\partial t}` is the time derivative of the electric displacement
`\displaystyle \mathbf{D}`.

In this problem, there is no current density (
`\displaystyle \mathbf{J} =0`) and no time-varying electric displacement (
`\displaystyle \frac{\partial \mathbf{D}}{\partial t} =0`). Therefore, the equation simplifies to:

`\displaystyle \\abla * \mathbf{H} =0`.

Taking the curl of the given magnetic field intensity
`\displaystyle \mathbf{R} =2\cos( 10^(10) t-600x)\hat{a}_(z)\, \text{Am}`:

`\displaystyle \\abla * \mathbf{R} =\\abla * ( 2\cos( 10^(10) t-600x)\hat{a}_(z)) \, \text{Am}`.

Using the curl identity and applying the chain rule, we can expand the expression:

`\displaystyle \\abla * \mathbf{R} =\left( \frac{\partial ( 2\cos( 10^(10) t-600x)) \hat{a}_(z)}{\partial y} -\frac{\partial ( 2\cos( 10^(10) t-600x)) \hat{a}_(z)}{\partial z}\right) \mathrm{d} x\mathrm{d} y\mathrm{d} z`.

Since the magnetic field intensity
`\displaystyle \mathbf{R}` is not dependent on
`\displaystyle y` or
`\displaystyle z`, the partial derivatives with respect to
`\displaystyle y` and
`\displaystyle z` are zero. Therefore, the expression further simplifies to:

`\displaystyle \\abla * \mathbf{R} =-\frac{\partial ( 2\cos( 10^(10) t-600x)) \hat{a}_(z)}{\partial x} \mathrm{d} x\mathrm{d} y\mathrm{d} z`.

Differentiating the cosine function with respect to
`\displaystyle x`:

`\displaystyle \\abla * \mathbf{R} =-2( 10^(10)) \sin( 10^(10) t-600x)\hat{a}_(z) \mathrm{d} x\mathrm{d} y\mathrm{d} z`.

Setting this expression equal to zero according to
`\displaystyle \\abla * \mathbf{H} =0`:

`\displaystyle -2( 10^(10)) \sin( 10^(10) t-600x)\hat{a}_(z) \mathrm{d} x\mathrm{d} y\mathrm{d} z =0`.

Since the equation should hold for any arbitrary values of
`\displaystyle \mathrm{d} x`,
`\displaystyle \mathrm{d} y`, and
`\displaystyle \mathrm{d} z`, we can equate the coefficient of each term to zero:

`\displaystyle -2( 10^(10)) \sin( 10^(10) t-600x) =0`.

Simplifying the equation:

`\displaystyle \sin( 10^(10) t-600x) =0`.

The sine function is equal to zero at certain values of
`\displaystyle ( 10^(10) t-600x)`:

`\displaystyle 10^(10) t-600x =n\pi`,

where
`\displaystyle n` is an integer. Rearranging the equation:

`\displaystyle x =( 10^(10) t-n\pi )/(600)`.

The equation provides a relationship between
`\displaystyle x` and
`\displaystyle t`, indicating that the magnetic field intensity is constant along lines of constant
`\displaystyle x` and
`\displaystyle t`. Therefore, the magnetic field intensity is uniform in the given medium.

Since the magnetic flux density
`\displaystyle B` is related to the magnetic field intensity
`\displaystyle H` through the equation
`\displaystyle B =\mu H`, where
`\displaystyle \mu` is the permeability of the medium, we can conclude that the magnetic flux density is also uniform in the medium.

Thus, the correct expression for the magnetic flux density in the given medium is:

`\displaystyle B =6\cos( 10^(10) t-600x)\hat{a}_(z)`.